Partial fractions integral problem

[suspenc3]

Hi, IM trying to evaluate this, and I can't get started..I tried integration by partial fractions and substitution but I keep getting stuck.

$$\int_0^2 \frac{x-3}{2x-3}dx$$

Any hints would help, Thanks

 

[StatusX]

It should be straightforward using partial fractions. Remember, since the polynomial in the numerator has the same degree as the one in the denominator, you need to add an extra unknown constant:

$$\frac{x-3}{2x-3} = \frac{A}{2x-3} +B$$

Or, by inspection, just write x-3=(x-3/2)-3/2=1/2(2x-3)-3/2.

 

[durt]

Consider that $$\frac{x-3}{2x-3}$$ has a vertical asymptote.

 

[benorin]

You can also do long (or synthetic) division to get

$$\frac{x-3}{2x-3}=\frac{1}{2}-\frac{3}{2(2x-3)}$$

 

[suspenc3]

oh ok thanks..I had part of that earlier..but the last step makes it work

Thanks

 

[suspenc3]

yes I got $$\frac{x-3}{2x-3}=\frac{1}{2}-\frac{3}{2(2x-3)}$$

..and continue to get $$-3ln(4x-6) + \frac{x}{2} \right]_2^0$$

=$$[-3ln2 +1] - [-3ln|-6|]$$

 

[d_leet]

yes I got $$\frac{x-3}{2x-3}=\frac{1}{2}-\frac{3}{2(2x-3)}$$

..and continue to get $$-3ln(4x-6) + \frac{x}{2} \right]_2^0$$

=$$[-3ln2 +1] - [-3ln|-6|]$$

I think that what durt mentioned should be reemphasized that this is not just a definite integral but an improper integral since the function you are integrating has a vertical assymptote at x = 2/3 which lies inside of the interval you are integrating over.

 

[Robokapp]

But it's not continuous as D_leet said so...I'd break it in two integrals...

remember that integral from a to b + inegral from b to c = integral from a to c. Because of the asymptote, it has a sign change in there also...it goes from + to - so that complicates the one-integral process.